Problem. Given four real numbers$a_1 \ge a_2 \ge a_3 \ge a_4 > 0$such that $a_1 + a_2 + a_3 + a_4 = 9$and $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 21$, find the minimum and maximum of $\frac{a_1}{a_4}$.
(Note: $a_1 + a_2 + a_3 + a_4 = 9$and $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 21$ along with $a_1, a_2, a_3, a_4\ge 0$ensure $a_i \ge 3/2, \forall i$. Thus, $\frac{a_1}{a_4}$ is bounded.)
Edit From @zetko's answer, I realize that we just ignore the boundary and find the points satisfying the Lagrange Multipliers equation system.So I ask a bad question :).
By the way, this is a special case of this question.
Discussion.
- Minimum
The minimum of $\frac{a_1}{a_4}$ is $\frac{14}{13} + \frac{3}{13}\sqrt{3}$ when$$a_1 = a_2 = \frac94 + \frac14\sqrt{3}, \quada_3 = a_4 = \frac94 - \frac14\sqrt{3}. \tag{1}$$
We may ask can we apply the Lagrange Multipliers by ignoring the constraints $a_1 \ge a_2 \ge a_3 \ge a_4$? In other words, are there $\lambda$ and $\mu$ such that$$\frac{\partial f}{\partial a_i} = 0, \quad i = 1, 2, 3, 4? \tag{2}$$where$$f(a_1, a_2, a_3, a_4) := \ln\frac{a_1}{a_4} - \lambda (a_1 + a_2 + a_3 + a_4 - 9) - \mu (a_1^2 + a_2^2 + a_3^2 + a_4^2 - 21).$$(Note: Take logarithm of the ratioin order to simplify the calculations.)
The answer is No. If we plug the minimizer (1) into (2), there are no such $\lambda$ and $\mu$.
We should apply the KKT conditions instead. We have$$\frac{\partial g}{\partial a_i} = 0, \quad i = 1, 2, 3, 4 \tag{3}$$where\begin{align*}g(a_1, a_2, a_3, a_4)&:= \ln\frac{a_1}{a_4} - \lambda (a_1 + a_2 + a_3 + a_4 - 9) - \mu (a_1^2 + a_2^2 + a_3^2 + a_4^2 - 21)\\&\qquad - r_1 (a_1 - a_2) - r_2(a_2 - a_3) - r_3(a_3 - a_4). \end{align*}For example, for the minimizer (1), we have$$\lambda = -\frac{28}{39}\sqrt{3},\mu = \frac{2}{13}\sqrt{3},r_1 = \frac{3}{13} - \frac{1}{39}\sqrt{3},r_2 = 0, r_3 = \frac{3}{13} + \frac{1}{39}\sqrt{3}.$$
$\phantom{2}$
- Maximum
The maximum is $\frac76 + \frac16\sqrt{13}$when$$a_1 = \frac{13}{6} + \frac16\sqrt{13}, \quad a_2 = a_3 = \frac73, \quad a_4 = \frac{13}{6} + \frac16\sqrt{13}. \tag{4}$$
Similarly, we may ask can we apply the Lagrange Multipliers by ignoring the constraints $a_1 \ge a_2 \ge a_3 \ge a_4$?This time, there are $\lambda$ and $\mu$ along with the maximizer (4) such that$$\frac{\partial f}{\partial a_i} = 0, \quad i = 1, 2, 3, 4? \tag{5}$$where$$f(a_1, a_2, a_3, a_4) := \ln\frac{a_1}{a_4} - \lambda (a_1 + a_2 + a_3 + a_4 - 9) - \mu (a_1^2 + a_2^2 + a_3^2 + a_4^2 - 21).$$
$\phantom{2}$
Question. For an optimization problem, if we apply the Lagrange Multipliers by ignoring the constraints $a_1 \ge a_2 \ge a_3 \ge a_4$ and there are such $\lambda$ and $\mu$ and a feasible $(a_1, a_2, a_3, a_4)$. Does it mean the KKT conditions is reduced to the Lagrange Multipliers?In other words, does it mean the KKT conditions does not have another KKT point $\lambda$ and $\mu$ and a feasible $(a_1, a_2, a_3, a_4)$?Or, we should apply the KKT conditions rather than the Lagrange Multipliers. We need to check all the KKT points.